Simplify and expand the following expression: $ \dfrac{1}{q + 8}- \dfrac{4}{3q - 27}+ \dfrac{3}{q^2 - q - 72} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3q - 27} = \dfrac{4}{3(q - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 - q - 72} = \dfrac{3}{(q + 8)(q - 9)}$ Now we have: $ \dfrac{1}{q + 8}- \dfrac{4}{3(q - 9)}+ \dfrac{3}{(q + 8)(q - 9)} $ The least common multiple of the denominators is: $ (q + 8)(q - 9)$ In order to get the first term over $(q + 8)(q - 9)$ , multiply by $\dfrac{3(q - 9)}{3(q - 9)}$ $ \dfrac{1}{q + 8} \times \dfrac{3(q - 9)}{3(q - 9)} = \dfrac{3(q - 9)}{(q + 8)(q - 9)} $ In order to get the second term over $(q + 8)(q - 9)$ , multiply by $\dfrac{q + 8}{q + 8}$ $ \dfrac{4}{3(q - 9)} \times \dfrac{q + 8}{q + 8} = \dfrac{4(q + 8)}{(q + 8)(q - 9)} $ In order to get the third term over $(q + 8)(q - 9)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{3}{(q + 8)(q - 9)} \times \dfrac{3}{3} = \dfrac{9}{(q + 8)(q - 9)} $ Now we have: $ \dfrac{3(q - 9)}{(q + 8)(q - 9)} - \dfrac{4(q + 8)}{(q + 8)(q - 9)} + \dfrac{9}{(q + 8)(q - 9)} $ $ = \dfrac{ 3(q - 9) - 4(q + 8) + 9} {(q + 8)(q - 9)} $ Expand: $ = \dfrac{3q - 27 - 4q - 32 + 9}{3q^2 - 3q - 216} $ $ = \dfrac{-q - 50}{3q^2 - 3q - 216}$